If I have a VSWR of 2:1 and a phase difference of 30 degrees between the forward and reflected signals how do I plot this on a Smith chart? For VSWR I use the circles radiating out from the center. Assuming an inductive load (reflected power leading), I have two alternatives for the phase line on the top of the chart, pointing towards the short circuit side or pointing towards the open circuit side.

There should only be one point on the SWR = 2 curve with a +30 degreephase shift.

Phase shift is given by the azimuth angle from the center of the chart to impedance, and is marked around the perimeter of the chart with 0 degreesat the bottom, +90 degrees in the middle of the right side, and +/- 180 degrees at the top. +30 degrees is in the lower right-hand corner.

Note: When a resistive load impedance is less than Z0, the reflected voltage undergoes a 180 degree phase shift. When a resistive load impedance is greater than Z0, the reflected current undergoes a 180 degree phase shift.

The first impedance has a reflection coefficient of 0.3333 at 150 degrees. That would put the two waves 150 degrees out of phase at the load, not 30 degrees out of phase.

The second impedance has a reflection coefficient of 0.3333 at 30 degrees. Seems that is the only load impedance for Z0=50 ohms that will cause an SWR=2:1 with Vref leading Vfor by 30 degrees. Note this is the 1.65+j0.6 impedance that I mentioned in my previous posting. When you un-normalize that impedance to 50 ohms you get the second impedance above.

...Each of the following mixed impedances gives a line from the center of a Smith chart with a thirty degree angle to the horizontal...

On my Smith Chart at least, the phase shift is measured starting with zeroat the bottom of the chart. +30 degrees is in the lower right quadrant.30 degrees from vertical at the top of the chart is a phase shift of+150 degrees, not +30 degrees. A horizontal line should have a phase shiftof +/- 90 degrees.

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